M4maths Previous Puzzles

28 Dec 2016 Permutation

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 0 which are divisible by 5 and none of the digits is repeated?

OPtion

1) 16
2) 18
3) 20
4) 24
5) 26
6) 28
7) 30
8) 32
9) 34
10)36

Solution

To be divisible by 5 unit digit of number should be either 5 or 0
We have total 6 digits out of which we cannot have 0 at the hundred’s place (else it will become a 2 digit number)
Unit place is filled with 5
a. Tens place can be filled with 0
Hundreds place can now be filled with (2, 3, 6, 7). We have 4 ways
So we have 1*1*4
b. Tens place can be filled with any of the digits (2, 3, 6, 7). We have 4 ways
Hundreds place can now be filled with remaining of 3 digits. We have 3 ways
So we have 1*4*3

Unit place is filled with 0
Tens place can be filled with any of the digits (2, 3, 5, 6, 7). We have 5 ways
Hundreds place can now be filled with remaining of 3 digits. We have 4 ways
So we have 1*5*4

So Total number of ways = 1*1*4 + 1*4*3 + 1*5*4 = 36

Option 10)

Correct Option: 10 Best Solution (14)

27 Dec 2016 Probability

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is greater than 15?

OPtion

1) 7/36
2) 8/36
3) 9/36
4) 5/36
5) 10/36
6) 11/36
7) 1/3
8) 13/36
9) 14/36
10)15/36

Solution

Total Number of possible outcomes = 6C1*6C1 = 36
Positive outcomes must satisfy D1*D2> 15
For D1 or D2 = 1 and 2 there is no possibility of getting D1*D2 > 15
For D1 = 3, we have D2=5 and 6 which satisfies D1*D2 > 15
(3,5) and (3,6)
For D1 = 4, we have D2=4, 5 and 6 which satisfies D1*D2 > 15
(4,4), (4,5) and (4,6)
For D1 = 5, we have D2=4, 5 and 6 which satisfies D1*D2 > 15
(5,4), (5,5) and (5,6)
For D1 = 6, we have D2=3, 4, 5 and 6 which satisfies D1*D2 > 15
(6,3), (6,4), (6,5) and (6,6)

Count of positive outcome = 11
Probability = Positive outcome/Total Possible Outcomes = 11/36
Option 6)

Correct Option: 6 Best Solution (18)

26 Dec 2016 Sequence

Find the next term
11, 13, 41, 58, .....?

OPtion

1) 63
2) 67
3) 73
4) 76
5) 86
6) 98
7) 125
8) 328
9) 595
10)625

Solution

Explanation:
11 + 1^2 + 1^3 = 13
13 + 1^2 + 3^3 = 41
41 + 4^2 + 1^3 = 58
Similarly
58 + 5^2 + 8^3 = 595

Option 9)

Correct Option: 9 Best Solution (2)

23 Dec 2016 Ages

Sum of the ages of a boy and his father is 70 years. Before 20 years Father's age was 9 times the age of the boy. What was the age of father in years at the time of birth of the child?

OPtion

1) 23
2) 24
3) 26
4) 27
5) 29
6) 33
7) 34
8) 36
9) 47
10)48

Solution

Let the present age of father and son be x and y respectively
According to the given condition
x+y = 70 ---- (i)
before 20 years
(x-20) = 9(y-20)
Replacing value of x from equation (i)
so 70-y-20 = 9(y-20)
=> 50-y = 9y-180
=> 10y = 230
=> y = 23 and x = 47
so at the time of birth of child, age of father = 47-23 = 24 years

Option 2)

Correct Option: 2 Best Solution (19)

22 Dec 2016 Probability

Cricket Team of Nepal needs to score 7 runs in last 2 balls to win the match. What is the probability that Nepal Team wins the match? Assume that:
- There were no wide balls or no balls bowled in these 2 balls.
- No Wicket has fallen in these 2 balls
- No Overthrows or extra runs were scored in these 2 balls
- In case if run(s)are scored on any of these 2 balls it can only be one of these values: 1, 2, 3, 4 & 6
This also implies that it’s quite possible that zero runs are scored on a particular ball

OPtion

1) 1/7
2) 1/6
3) 1/5
4) 1/4
5) 1/3
6) 2/7
7) 2/5
8) 2/3
9) 3/5
10)3/4

Solution

On any of these 2 balls the possible runs which can be scored are: 0,1,2,3,4,6
So total possible number of runs which can be scored in 2 balls are = 6C1*6C1 = 36
Let’sassume that run scored on first ball is x and 2nd ball is y
To win the match x+y>= 7. Possible pair of values of x and y are:
(x, y)
(1, 6), (6, 1), (2, 6), (6, 2), (3, 4), (4, 3), (3, 6), (6, 3), (4, 4), (4, 6), (6, 4), (6, 6)
Count of set of values for Nepal to win the match = 12
Probability = Positive outcome/Total Possible Outcomes = 12/36 = 1/3

Correct Option: 5 Best Solution (7)

21 Dec 2016 Coding Decoding

If in coded language
CHINA = 107
JAPAN = 134
NEPAL = 136
Then
INDIA = ?

OPtion

1) 149
2) 143
3) 138
4) 126
5) 114
6) 108
7) 100
8) 90
9) 84
10)78

Solution

A=1, B=2, C=3, D=4, E=5, F=6, G=7, H=8, I=9......
Each string is coded as: Sum of (Position of each letter in string*Position of each letter kin alphabet)
CHINA = 1*3+2*8+3*9+4*14+5*1 = 107
JAPAN = 1*10+2*1+3*16+4*1+5*14 = 134
NEPAL = 1*14+2*5+3*16+4*1+5*12 = 136
INDIA = 1*9+2*14+3*4+4*9+5*1 = 90

Option 8)

Correct Option: 8 Best Solution (1)

20 Dec 2016 Stocks

Jennifer invests a part of Rs. 27,000 in 12% stock at Rs. 120 and the remainder in 15% stock at Rs. 135. If her total dividend per annum is Rs. 2828, how much does she invest in 12% stock at Rs. 120?

OPtion

1) Rs. 4000
2) Rs. 7640
3) Rs. 9480
4) Rs. 11520
5) Rs. 12840
6) Rs. 15480
7) Rs. 16960
8) Rs. 18580
9) Rs. 20480
10)Rs. 22560

Solution

Let investment in 12% stock be Rs. x.
Then, investment in 15% stock = Rs. (27000 - x).
Therefore (12/120)*x + (15/135)(27000 - x) = 2828
=> x/10 + 1/9(27000-x) = 2828
=> x/10 + 3000 - x/9 = 2828
=> 9x + 3000*90 - 10x = 2828*90
=> x = 15480

Option 6)

Correct Option: 6 Best Solution (6)

19 Dec 2016 Probability

What is the probability of getting a sum 12 from three throws of a dice?

OPtion

1) 11/108
2) 7/72
3) 1/9
4) 25/216
5) 13/108
6) 1/8
7) 7/54
8) 29/216
9) 5/36
10)31/216

Solution

Total Number of possible outcomes = 6C1*6C1*6C1 = 6*6*6 = 216
Positive outcomes must satisfy D1+D2+D3= 12
(1,5,6) (1,6,5) (2,4,6) (2,6,4) (2,5,5) (3,3,6) (3,6,3) (3,4,5) (3,5,4) (4,2,6) (4,6,2) (4,3,5) (4,5,3) (4,4,4) (5,1,6) (5,6,1) (5,2,5) (5,5,2) (5,3,4) (5,4,3) (6,1,5) (6,5,1) (6,2,4) (6,4,2) (6,3,3)
Count of positive outcome = 25
Probability = Positive outcome/Total Possible Outcomes = 25/216

Option 4)

Correct Option: 4 Best Solution (18)

16 Dec 2016 Combination

The Indian Cricket team consists of 16 players. It includes 1 wicket keeper and 7 bowlers. In how many ways can a playing eleven be selected if we have to select 1 wicket keeper and at least 5 bowlers?

OPtion

1) 496
2) 784
3) 932
4) 1176
5) 1284
6) 1336
7) 1492
8) 1582
9) 1666
10)1722

Solution

We have to choose 11 players including:
- 1 wicket keeper and 5 bowlers or
- 1 wicket keeper and 6 bowlers or
- 1 wicket keeper and 7 bowlers.
Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players
=1C1*7C5*8C5= 1*21*56 = 1176
Number of ways of selecting 1 wicket keeper, 6 bowlers and 4 other players
=1C1*7C6*8C4= 1*7*70 = 490
Number of ways of selecting 1 wicket keeper, 7 bowlers and 3 other players
=1C1*7C7*8C3=1*1*56 = 56
=> Total number of ways of selecting the team:
=1176+490+56= 1722

Option 10)

Correct Option: 10 Best Solution (0)

15 Dec 2016 Digits

When you reverse the digits of the number 14 the number increases by 27. How many other two digit numbers increase by 27 when their digits are reversed?

OPtion

1) 1
2) 2
3) 3
4) 4
5) 5
6) 6
7) 7
8) 8
9) 0
10)None of these

Solution

10x + y - (10y + x ) = 27
=> 9x-9y = 27
=> x - y = 3
Following numbers satisfy the criteria of x-y=3
14, 25, 36, 47, 58, 69
So there are 4 other 2 digit numbers apart from 14 which increase by 27 when their digits are reversed

Option 5)

Correct Option: 5 Best Solution (0)